AME 451

Project Report

Submission Date: August 13, 2014

Instructor: Mukherjee, Jyoti

Team members: Molani, Mohammad Salem

Jun, Youra

Cao, Zhe

Table of ContentsTable of Contents........................................................................................................................................2

Introduction.................................................................................................................................................2

Results and calculations..............................................................................................................................2

Part 1. Vehicle information and measurements.......................................................................................2

Part 2. Homework Problems....................................................................................................................4

HW 1A................................................................................................................................................4

HW 2...................................................................................................................................................7

HW3..................................................................................................................................................10

Conclusion.................................................................................................................................................19

IntroductionHow would it be if the students get to work and apply the homework problems calculations into a real car? This project is one of the best opportunities for the group members in order to face the real engineering duty. The group started from identifying the vehicle information and gets to know about the vehicle features and how things work. After manually measuring the data needed from the car to solve the homework problems, the group came up with reasonable results beside that, there are numbers assumed by the group members for solving the problems.

Results and calculations

Part 1. Vehicle information and measurements

Vehicle Used : 2004 Mitsubishi Galant 3.8 liter GTS

Engine ID: 6G75

Vehicle ID (VIN): 4A3AB76S74E106239

Tire ID: DOT 93 ENB0621311

Wheelbase (L): 108.5”

Tread Width: 8.5”

Static Loaded Radius(R): 13”

When Car is empty;

Gross Vehicle Weight : 4541 lbs

Static Front Wheel Load(Rf) : 2491 lbs

Static Rare Wheel Load(Rr) : 2050 lbs

Distance from the ground to the bottom of the vehicle (hi) : 8.75”

Vertical location of the center of gravity (ycg) : 24” from the ground

Horizontal location of the center of gravity (xcg) : 49” from the front wheel

o Calculating the horizontal location of the center of gravity

∑T f=4541lbs ∙ xcg−2050 lbs∙108.3∈¿=0 ¿

x=2050 lbs∙108.3∈ ¿4541lbs

=49∈¿¿

When there are two passengers;

Weight of two passengers (Wp) : 398.7 lbs

Distance from the ground to the bottom of the vehicle (hf) : 7.5”

Vertical location of the center of gravity of the passengers (yp) : 29” from the ground

Horizontal location of the center of gravity of the passengers(xp) : 19” from the front

wheel.

Spring rate (k):

o Calculating the spring rates

k=W2δ

=W p

2(hi−h f )=

W p

2(hi−h f )=398.7 lbs

2¿¿

Brake Gain of the front wheel (Gf) : 0.22 N ∙mkPa per wheel

Brake Gain of the rear wheel (Gr) : 0.15 N ∙mkPa per wheel

Part 2. Homework Problems

HW 1A

Using the data found in part 1, answer the following questions.

a) Find the longitudinal position of the combined CG.

x=∑ (weight ∙ X )

∑ weight=4541 lbs¿¿

b) Determine the axle loads of the loaded vehicle.

Balancing the moment about front axle.

Rr ∙108.5= (4541lbs+398.7lbs)(52.3 ¿

Rr=2381.07 lbs

Balancing vertical direction forces

Rr+R f=4541 lbs+398.7lb

R f=2558.63 lbs

c) Find the vertical position of the combined CG.

y=∑ (weight ∙ y )

∑ weight=4541 lbs¿¿

d) What are the axle loads when it accelerates at 3m/sec2?

Forward acting force

F=ma=( 4541lbs32.2 ftsec2 )( 3m

sec2 )( 3.28 ft1m )=1387.68 lbs

Balancing moments about the front axle

(1387.68 lbs)(24.4 )+ {R} rsub {r } ¿

Rr=2069 lbs

Balancing moments about the rear axle

(1387.68 lbs)(24.4 )− {R} rsub { f } (108.5)=(4541lbs+398.7lbs)(108.5−52.3 )

Rr=2870.69 lbs

e) What are the axle loads when the trailer shown in the figure is connected up? The trailer

hitch on the car is 90mm behind the rear axle.

Balancing moments about hitch

610 kg ( 9.81m / s2 ) (2.8m )=Rt(3.25m)

Rt=5155.5 N=1159.0 lbs

Balancing the force in order to find the force at hitch

Rh=610kg(9.81m /s2)−5155.5N=828.6 N=186.28 lbs

Taking the car and balancing moment about front axle

Rr ¿

Rr=2633.59 lbs

Balancing vertical direction forces

Rr+R f=4541 lbs+398.7lb+186.28 lbs

R f=2492.39 lbs

f) Because of the high rear axle load on the car, the owner installs a “load-equalizing” hitch.

This hitch puts a torque of 1500Nm (=13276.1lbs*in) into the connection point as

illustrated in the figure below. What are the wheel loads (car and trailer) with the load

equalizer?

Balancing the moment about the hitch.

610 kg ( 9.81m / s2 ) (2.8m )+1500 N ∙m=Rt(3.25m)

Rt=5617 N=1262.75 lbs

load at hitch=Rh=610 kg (9.81m /s2)−5617 N=367.1 N=82.53lbs

Taking the car and balancing moment about front axle

Rr (108.5 )+13276.1lbs∙in=(4541lbs+398.7lbs)(52.3)+82.53lbs(108.5 +38.57)

Rr=2370.58 lbs

HW 2

There are 3 problems in the HW 2. We are going to use all data we measured.

Q1. If the vehicle is running at 80km/hr up a 2% grade. The drag from aerodynamics and

rolling resistance is 572N. It is operating in third gear, which has a ratio of 1.2, and has a

final drive ratio of 3.5.With the given data and the measurements in part 1, answer the

following questions.

(a) What is the engine speed (RPM)?

Angular velocity = v

2∗π∗r=

80 kmhr

∗10.936 ¿s

2∗π∗13∈¿=422.86 rads

¿

n1n2

=t 1t 2

, w2w1

= t 1t 2

422.86w1

=1.21 ,w1=349.47 rad /s

(b) How much torque (N-m) does the engine develop to maintain a constant speed if the

driveline efficiency is 91%.

4541 lb = 2059.88 kg = 20207.45 N; 13 in = 0.33 m

Ft = 572 N + 0.02 * 20207.45 N = 976.15 N

Tt = Ft * r = 976.15 N * 0.33 m = 322.13 Nm

T2 = Tt

Nr∗Nf∗η= 322.13Nm

1.21∗3.5∗0.91=83.59 Nm

(c) If the engine has a brake-specific fuel consumption of 0.3 kg/kw-h at these conditions,

what is the fuel consumption (kg/km)? Also give it in liters/100 km, assuming gasoline

weighs 0.78 kg/liter.

Power developed by the engine = T2 * w1 = 83.59 Nm * 349.47 rad/s = 29.21 KW

Fuel consumption = 29.21 KW * 0.3 kg /hrkw

=8.76 kghr

=8.76kg/hr80 km /hr

=0.11 kgkm

= 0.11kgkm

∗1 liter

0.78 kg=0.14 liter

km = 14

liters100 km

Q2. The engine of a vehicle has a moment of inertia of 0.277 kg-m2. It is operating in first

gear (ratio of 7.53), has a final drive ratio of 4.11, and the tire radius is 0.508m.

(a) What is its effective mass of the engine?

M eff = I∗N2

r2 =0.277 kg−m2∗(7.53∗4.11)2

(0.33m)2 =2436.27 kg

(b) What is the “mass factor” for this condition?

Mass Factor = M+M eff

M=2059.88 kg+2436.27 kg

2059.88 kg=2.18

Q3. The vehicle can accelerate to 300mph in a quarter mile. The performance can be

modeled rather simply by assuming that the engine delivers constant power during

acceleration, and neglecting rolling resistance and aerodynamic drag forces.

(a) Derive and solve the differential equation relating velocity to distance for this case.

F x=M∗ax=M∗dV

dt

P=F x∗V ;F x=PV;∧V=dx

dt

PV

∗dt=

PV

∗dx

V=M∗dV

∫0

x

dx=MP

∗∫0

V

V 2∗dV

X=

MP

∗V 3

3

(b) If the dragster weighs 4541 lb, how much power is develop to achieve acceleration to 300

mph in the quarter mile? (Neglect driveline inertia)

X=52804

=1320 ft

300 mph=440 ftsec

P=

MX

∗V 3

3=

4541lb

32.2 ftsec2∗1320 ft

∗(440 ftsec )

3

3=3,033,601 ft−lb

sec

= 5515.42 HP

(c) How much time is required to cover the quarter mile?

∫0

T

dt=MP ∫

0

V

V∗dV

T=MP

V 2

2= 4541 lb

32.2 ftsec 2∗3,033,601 ft−lb

sec

(440 ftsec

)2

2=4.5 sec

HW3

Q. You have been asked to help design the brake system on a new light truck. The truck

must meet FMVSS 105 performance requirements for deceleration (take this to mean 20

ft/sec2 lightly loaded, and 19 ft/sec2 fully loaded on an 81 Skid Number surface), as well

as corporate requirements for 0.25 g braking performance on a 30 Skid Number surface.

The important specifications of the vehicle are:

L=108.5” Rtire=13”

Brake gains (per wheel) 0.22 Nm/kPa (front) 0.15Nm/kPa (rear)

lightly-loaded fully-loaded

CG height 24" 24.4"

front axle weight 2491lbs 2558.63lbs

rear axle weight 2050lbs 2381.07lbs

You are to select a pressure proportioning system (break pressure and rise rate) and then

determine the efficiency of the system for the lightly-loaded and fully-loaded conditions. Do this

via the following steps:

(a) Calculate and plot the four performance triangles on the front/ rear brake force diagram

for all conditions. (Show your calculations for the intercept points.)

First step is to find the maximum front and rear brake forces for each condition.

F xfm=μp(W fs+

hLF❑xr)

1−μphL

o Front brake force axis intercepts:

Lightly-loaded & 81 SN, F xfm=0.81(2491 lbs)

1−0.81(24 )} over {(108.5)=2458.1 lbs

Fully-loaded & 81 SN, F xfm=0.81(2558.63 lbs)

1−0.81(24.4 )} over {(108.5)=2534.09 lbs

Lightly-loaded & 30 SN, F xfm=0.3(2491 lbs )

1−0.3(24 )} over {(108.5)=800.42 lbs

Fully- loaded & 30SN, F xfm=0.3(2558.63 lbs)

1−0.3(24.4 )} over {(108.5)=823.12 lbs

F xrm=μp(W fs−

hLF❑xr)

1+μphL

o Rear brake force axis intercepts:

Lightly-loaded & 81 SN, F xfm=0.81(2050 lbs)

1+0.81(24 )} over {(108.5)=1408.19lbs

Fully-loaded & 81 SN, F xfm=0.81(2381.07 lbs)

1+0.81(24.4 )} over {(108.5)=1631.48lbs

Lightly-loaded & 30 SN, F xfm=0.3(2050 lbs)

1+0.3(24 )} over {(108.5)=576.73lbs

Fully- loaded & 30SN, F xfm=0.3(2381.07 lbs)

1+0.3(24.4 )} over {(108.5)=669.17 lbs

Slope of front axle

m=

μphL

1−μphL

Lightly-loaded & 81 SN, m=0.81¿¿¿

Fully-loaded & 81 SN, m=0.81 ¿¿¿

Lightly-loaded & 30 SN, m=0.3 ¿¿¿

Fully- loaded & 30SN, m=0.3 ¿¿¿

Slope of rear axle

m=

−μphL

1+μ phL

Lightly-loaded & 81 SN, m=−0.81¿¿¿

Fully-loaded & 81 SN, m=−0.81¿¿¿

Lightly-loaded & 30 SN, m=−0.3¿¿¿

Fully- loaded & 30SN, m=−0.3¿¿¿

Intersection point

F xf i=μp(W fs+μ pW ( hL ))F xr i=μp(W rs−μ pW ( hL ))Lightly-loaded & 81 SN,

F xf i=0.81 (2491 lbs+0.81 (4541 lbs ) (24 } over {108.5 ) )=2676.74 lbs

F xr i=0.81 (2050lbs−0.81 (4541 lbs ) (24 } over {108.5 ) )=1001.47lbs

Fully-loaded & 81 SN,

F xf i=0.81 (2558.63lbs+0.81 ( 4939.7lbs ) (24.4 } over {108.5 ) )=2801.33lbs

F xr i=0.81 (2381.07 lbs−0.81 (4939.7 lbs ) (24.4} over {108.5 ) )=1199.83 lbs

Lightly-loaded & 30 SN,

F xf i=0.3 (2491lbs+0.3 ( 4541 lbs ) (24 } over {108.5 ) )=837.70lbs

F xr i=0.3 (2050 lbs−0.3 (4541 lbs ) (24 } over {108.5 ) )=524.6 lbs

Fully- loaded & 30SN,

F xf i=0.3 (2558.63 lbs+0.3 (4939.7 lbs ) (24.4 } over {108.5) )=867.57 lbs

F xr i=0.3 (2381.07 lbs−0.3 ( 4939.7lbs ) (24.4 } over {108.5 ) )=614.34 lbs

Fxmf (lbs) Fxmr (lbs)slope for

front

slope for

rearFxfi (lbs) Fxri (lbs)

81SN,lightly

loaded2458.14 1408.19 0.2183 -0.1519 2676.74 1001.47

81SN,fully 2534.09 1631.48 0.2227 -0.1541 2801.33 1199.83

loaded

30SN,lightly

loaded800.42 576.73 0.0711 -0.0622 837.70 524.60

30SN,fully

loaded823.12 669.17 0.0723 -0.0632 867.57 614.34

Deceleration line

Lightly-loaded & 81 SN, F❑tot=ma= (2491lbs+2050 lbs )

32.2 ftsec2

× 20 ftsec2 =2820.5 lbs

Fully-loaded & 81 SN, F❑tot=ma= (2558.63lbs+2381.07 lbs )

32.2 ftsec2

× 19 ftsec2 =2914.73lbs

Lightly-loaded & 30 SN,

F❑tot=ma= (2491lbs+2050 lbs )32.2 ftsec2

×0.25 (32.2 ftsec2 )=1135.25 lbs

Fully-loaded& 30S,

F❑tot=ma= (2558.63 lbs+2381.07 lbs )32.2 ftsec2

×0.25( 32.2 ftsec2 )=1234.93 lbs

Using calculator, find the intersection between the deceleration line and the force line.

Lightly-loaded & 81 SN, (297.43, 2523.07) and (1155.27, 1665.26)

Fully-loaded & 81 SN, (1397.71, 1517.02) and (311.31, 2603.42)

Lightly-loaded & 30 SN, (539.69, 595.56) and (312.60, 822.65)

Fully- loaded & 30SN, (631.00, 603.93) and (384.04, 850.89)

(b) Select the parameters for a proportioning valve (pressure break point and slope) to be

used and show it on the diagram

The diagram below shows the performance triangles for this vehicle. Since the brake

gains were given be proportioning line must have an initial slope of 0.220.15

=1.467 . this is

slope is satisfactory for the lower triangles, but must change a higher slope to meet the

high lower triangles.

It would be reasonable to select the pressure break point for the pressure proportioning

valve equivalent to 4003.4 N (900 lb) of force on the front brakes. The associated

pressure is:

Pa=4003.4 N∗0.33m

2∗0.22N−m /Kpa=3002.55 kPa

At this pressure the rear brake force is:

F r=2∗0.15N− m

Kpa∗3002.55 Kpa

0.33m=2729.6N = 613.64 lbf

A reasonable pressure proportioning can be obtained by choosing the second segment of

the line to go to a rear brake force of 6227.51 N (1400 lb) when the front brakes go up to

13345 N (3000 lb). The associated brake pressures at this condition are:

Paf=13345 N∗0.33m

2∗0.22N−m /Kpa=10008.75Kpa

Par=6227.51N∗0.33m2∗0.15 N−m /Kpa

=6850.26 Kpa

The slope for the proportioning is:

S=∆Par

∆Paf= 6850.26−3002.55

10008.75−3002.55=0.549

The proportioning valve is denoted as 3002.55

0.549

%Graphing Front Brake Force Vs Rear Brake Force % For lightly loaded & 81 SN, fxr1=[1001.47 1408.19];fxf1=[2676.74 0]; plot(fxr1,fxf1); hold on fxr2=[0 1001.47];fxf2=[2458.14 2676.74]; plot(fxr2,fxf2); % For fully loaded & 81 SN, fxr3=[1199.83 1631.48];fxf3=[2801.33 0]; plot(fxr3,fxf3); fxr4=[0 1199.83];fxf4=[2534.09 2801.33]; plot(fxr4,fxf4); % For lightly loaded & 30 SN, fxr5=[524.60 576.73];fxf5=[837.70 0]; plot(fxr5,fxf5); fxr6=[0 524.60];fxf6=[800.42 837.70]; plot(fxr6,fxf6); % For fully loaded & 30 SN, fxr7=[614.34 669.17];fxf7=[867.57 0]; plot(fxr7,fxf7); fxr8=[0 614.34];fxf8=[823.12 867.57]; plot(fxr8,fxf8); %Drawing the deceleration line fxr9=[297.43 1155.27];fxf9=[2523.07 1665.26]; fxr10=[1397.71 311.31];fxf10=[1517.02 2603.42]; fxr11=[539.69 312.60];fxf11=[595.56 822.65]; fxr12=[631.00 384.04];fxf12=[603.93 850.89]; plot(fxr9,fxf9);plot(fxr10,fxf10);plot(fxr11,fxf11);plot(fxr12,fxf12); %Drawing the proportioning line fxr13=[0 700]; fxf13=[0 1.467*700]; fxr14=[700 1400]; fxf14=[1.467*700 3000]; plot(fxr13,fxf13); plot(fxr14,fxf14); xlabel('Rare Brake Force (lbs)'); ylabel('Front Brake Force (lbs)'); title('Brake Force Diagram'); hold off

(c) Calculate and plot the braking efficiency for the lightly-loaded and fully-loaded vehicle

as a function of application pressure for the design you selected.

Lightly loaded

Gf=0.22 Gr=0.15 R(tire)=0.33m Pa=3002.55kPa m(prop)=0.549

Wfs=2491lbs=11080.52N Wrs=2050lbs=9118.85N h=0.6096m WB=2.7559m

Pf(kPa) Pr(kPa) Fxf(N) Fxr(N) Dx(g) Wf(kg) Wr(kg) µf µr ηf ηr ηb

10 10 13.33 9.090.00

11130.0

2929.0

40.00

10.00

10.92

31.11

30.92

3

100 100 133.33 90.910.01

11134.5

7924.4

90.01

20.01

00.92

71.10

80.92

7

500 500 666.67 454.550.05

61154.7

9904.2

70.05

90.05

10.94

31.08

30.94

3

1000 1000 1333.33 909.090.11

11180.0

8878.9

80.11

50.10

50.96

41.05

30.96

4

2000 2000 2666.671818.1

80.22

21230.6

4828.4

20.22

10.22

41.00

50.99

20.99

2

3000 3000 4000.002727.2

70.33

31281.2

0777.8

60.31

80.35

71.04

60.93

20.93

2

40003501.2

8 5333.333182.9

80.42

21321.5

4737.5

20.41

10.44

01.02

50.95

80.95

8

50004001.2

8 6666.673637.5

20.51

01361.8

5697.2

10.49

90.53

21.02

20.95

90.95

9

60004501.2

8 8000.004092.0

70.59

91402.1

7656.8

90.58

20.63

51.02

90.94

30.94

3

70005001.2

8 9333.334546.6

10.68

71442.4

8616.5

80.66

00.75

21.04

20.91

40.91

4

80005501.2

810666.6

75001.1

60.77

61482.7

9576.2

60.73

30.88

51.05

80.87

70.87

7

90006001.2

812000.0

05455.7

00.86

41523.1

1535.9

50.80

31.03

81.07

60.83

30.83

3

100006501.2

813333.3

35910.2

50.95

31563.4

2495.6

40.86

91.21

61.09

60.78

40.78

4

110007001.2

814666.6

76364.8

01.04

11603.7

4455.3

20.93

21.42

51.11

70.73

10.73

1

Fully - loaded

Gf=0.22 Gr=0.15 R(tire)=0.33m Pa=3002.55kPa m(prop)=0.549

Wfs=2558.63lbs=11381.35N Wrs=2381.07lbs=10591.53N h=0.620m WB=2.7559m

Pf(kPa) Pr(kPa) Fxf(N) Fxr(N) Dx(g) Wf(kg) Wr(kg) µf µr ηf ηr ηb

10 10 13.33 9.090.00

11160.6

91079.1

50.00

10.00

10.87

21.18

80.87

2

100 100 133.33 90.910.01

01165.3

21074.5

20.01

20.00

90.87

51.18

30.87

5

500 500 666.67 454.550.05

11185.8

91053.9

50.05

70.04

40.89

01.16

10.89

0

1000 1000 1333.33 909.090.10

21211.6

01028.2

40.11

20.09

00.91

01.13

20.91

0

2000 2000 2666.671818.1

80.20

41263.0

3 976.820.21

50.19

00.94

81.07

60.94

8

3000 3000 4000.002727.2

70.30

61314.4

5 925.390.31

00.30

00.98

71.01

90.98

7

40003501.2

8 5333.333182.9

80.38

81355.4

8 884.360.40

10.36

70.96

61.05

60.96

6

50004001.2

8 6666.673637.5

20.46

91396.4

8 843.360.48

70.44

00.96

41.06

70.96

4

60004501.2

8 8000.004092.0

70.55

01437.4

8 802.360.56

70.52

00.97

01.05

90.97

0

70005001.2

8 9333.334546.6

10.63

21478.4

9 761.360.64

40.60

90.98

21.03

80.98

2

80005501.2

810666.6

75001.1

60.71

31519.4

9 720.360.71

60.70

80.99

61.00

80.99

6

90006001.2

812000.0

05455.7

00.79

41560.4

9 679.360.78

40.81

91.01

30.97

00.97

0

100006501.2

813333.3

35910.2

50.87

61601.4

9 638.360.84

90.94

41.03

20.92

80.92

8

110007001.2

814666.6

76364.8

00.95

71642.4

9 597.350.91

01.08

61.05

20.88

10.88

1

0 2000 4000 6000 8000 10000 120000.000

0.200

0.400

0.600

0.800

1.000

1.200

Efficiency

Lightly-LoadedMoving average (Lightly-Loaded)Fully-loadedMoving average (Fully-loaded)

Pressure (kPa)

Efficie

ncy

ConclusionThe following major assumptions that the team has made so far where included in the homework

problems and applied resulting not actual results:

Gross Vehicle Weight

Static Front Wheel Load(Rf)

Static Rare Wheel Load(Rr)

The Break gains, CG heights, and Horizontal location of passenger CG.

Last but not least, the project was one of the best ways to learn about real life vehicle. The group was

facing lots of problems. One of the problems that the group faced is identifying the vehicle, the

engine, and the tire ID. The group googled it and get the answer which states that the information

should be inside the hood behind the engine in a silver sticker. Also, measuring the value was one of

a kind experience to the group when trying to measure the lightly loaded and fully loaded.